Problem: What is the value of $\dfrac{d}{dx}\left(\dfrac{4-x^2}{3x-2}\right)$ at $x=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $-1$ (Choice C) C $2$ (Choice D) D $1$
Solution: Let's first find the expression for $\dfrac{d}{dx}\left(\dfrac{4-x^2}{3x-2}\right)$ (i.e. for any input value $x$ ). Then, we can plug $x=2$ and evaluate. $\dfrac{4-x^2}{3x-2}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d d x ( 4 − x 2 3 x − 2 ) = ( 3 x − 2 ) d d x ( 4 − x 2 ) − ( 4 − x 2 ) d d x ( 3 x − 2 ) ( 3 x − 2 ) 2 = ( 3 x − 2 ) ( − 2 x ) − ( 4 − x 2 ) ( 3 ) ( 3 x − 2 ) 2 = − 6 x 2 + 4 x − 12 + 3 x 2 ( 3 x − 2 ) 2 = − 3 x 2 + 4 x − 12 ( 3 x − 2 ) 2 The quotient rule Differentiate ( 4 − x 2 ) & ( 3 x − 2 ) Expand \begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{4-x^2}{3x-2}\right) \\\\ &=\dfrac{(3x-2)\dfrac{d}{dx}(4-x^2)-(4-x^2)\dfrac{d}{dx}(3x-2)}{(3x-2)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(3x-2)(-2x)-(4-x^2)(3)}{(3x-2)^2}&&\gray{\text{Differentiate }(4-x^2)\text{ & }(3x-2)} \\\\ &=\dfrac{-6x^2+4x-12+3x^2}{(3x-2)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{-3x^2+4x-12}{(3x-2)^2} \end{aligned} So we found that $\dfrac{d}{dx}\left(\dfrac{4-x^2}{3x-2}\right)=\dfrac{-3x^2+4x-12}{(3x-2)^2}$. Now let's plug $x= 2$ : $\begin{aligned} &\phantom{=}\dfrac{-3( 2)^2+4( 2)-12}{(3( 2)-2)^2} \\\\ &=\dfrac{-12+8-12}{4^2} \\\\ &=-1 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{4-x^2}{3x-2}\right)$ at $x=2$ is $-1$.